Olympiad Corner #1

Q1Q1 If SnS_{ n } is the sum

Sn=1+12+13+.........+1n(n>2),provethatn(n+1)1nn<Sn<n(n1)n1(n1)S_{ n }=1+\frac { 1 }{ 2 } +\frac { 1 }{ 3 } +.........+\frac { 1 }{ n } (n>2),\quad prove\quad that\\ n{ (n+1) }^{ \frac { 1 }{ n } }-n<{ S }_{ n }<n-(n-1){ n }^{ \frac { -1 }{ (n-1) } }

Q2Q2 If a1a_{1}, a2a_{2}, a3a_{3},........, ana_{n} are positive real numbers, prove that

i<jaiajn12(a1+a2+...+an)\sum _{ i<j }^{ }{ \sqrt { { a }_{ i }{ a }_{ j } } } \le \frac { n-1 }{ 2 } ({ a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n })

Note by Satyajit Ghosh
3 years, 9 months ago

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By using A.MG.MA.M-G.M we get

ω=2ωa1aω((n1)a1+ω=2ωaω)2\sum_{ω=2}^{ω}\sqrt{a_{1}a_{ω}}\leq(\frac{(n-1)a_{1}+\sum_{ω=2}^{ω}a_{ω})}{2}

ω=3ωa2aω((n2)a2+ω=3ωaω)2\sum_{ω=3}^{ω}\sqrt{a_{2}a_{ω}}\leq(\frac{(n-2)a_{2}+\sum_{ω=3}^{ω}a_{ω})}{2} . . . aω1aωaω1+aω2\sqrt{a_{ω-1}a_{ω}}\leq\frac{a_{ω-1}+a_{ω}}{2}

Adding all above equations we get

I<JaIaJ(n1)(ω=1ωaω)2\sum_{I<J}\sqrt{a_{I}a_{J}}\leq\frac{(n-1)( \sum_{ω=1}^{ω}a_{ω})}{2}

Hence proved !!!

Shivam Jadhav - 3 years, 9 months ago

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Thanks but I got it after the hint was dropped by dev. I wanted the hint for 1st question. Can you give me the hints or provide the solution?

Satyajit Ghosh - 3 years, 9 months ago

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2) Direct using AM GM

Dev Sharma - 3 years, 9 months ago

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Can you show?and what about 1st?

Satyajit Ghosh - 3 years, 9 months ago

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Apply am gm on a1&a2, a1&a3,...,a1&an,a2&a3........... Add them all

Dev Sharma - 3 years, 9 months ago

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@Dev Sharma What about the 1st one?

Satyajit Ghosh - 3 years, 9 months ago

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@Dev Sharma Thanks I got it

Satyajit Ghosh - 3 years, 9 months ago

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@Dev Sharma @Kushagra Sahni @Nihar Mahajan @Surya Prakash @Swapnil Das Adarsh Kumar

Satyajit Ghosh - 3 years, 9 months ago

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Did 2nd on my own

Kushagra Sahni - 3 years, 9 months ago

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Did you also use AM-GM?

Satyajit Ghosh - 3 years, 9 months ago

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@Satyajit Ghosh Yeah!!

Kushagra Sahni - 3 years, 9 months ago

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@Kushagra Sahni And first question?

Satyajit Ghosh - 3 years, 9 months ago

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